From the foot of an inclined plane, whose angle is tan (1/3), a shot is projected with a velocity of 10 m/s at an angle of 60° with the horizontal up the plane. Find the time of flight and range. (g = 9.8 m/s)
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Answer:20/3
Explanation:
theta= the angle of body from horizontal = 60°
Alpha= Angle of inclination 30°
The formula for range of a projectile on inclined plane is
V^2[sin(2theta-alpha) - sin(alpha)/gcos^2(alpha)]
[((10)^2×sin(120-30) - sin(30))/(9.8×3/2)]
Putting in the values we get-
(100×1 - 1/2)/4.9×3
Which will roughly be equal to 20/3
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